# EM waves and the direction of E-fields & B-Fields

1. May 5, 2013

### cryora

What I'm wondering is whether or not the E-fields and B-fields making up an EM wave always point outwards. Would it be possible to have an EM wave where either the E-fields or B-fields, or both, point inwards towards the path of propagation? The only diagrams I see in books and the internet alike either have them both point outward, or not have any indication of direction at all.

I understand that an EM wave consists of E-fields and B-fields changing in such a way that they are self-sustaining, and that this change doesn't necessarily have to be sinusoidal. However, I don't know exactly how this is done, such as what equations describes how changing E-fields create B-fields (and vice versa) in a straightforward manner. I know there are Maxwell Equations, but they don't directly tell you, for instance, the type of E-field created, when you have a given changing B-field, although that could be because I haven't studied those equations in enough depth to know how to apply them in such a way. Please forgive me as the only knowledge I have about EM waves is from a lower division General Physics course.

The reason why I wonder this is because of a question asked on Yahoo! Answers.
If it is true that the E-fields only point outwards from the path of an EM wave, wouldn't that break some sort of symmetry? I mean, I know symmetry isn't something to expect in everything, but wouldn't the fact that the E-fields only point outwards give the photon similar properties as a positively charged particle? So for instance, if you had positively charged particles arranged so that they are located within of the E-fields of an EM wave, wouldn't the EM wave cause those positively charged particles to be deflected away?

2. May 5, 2013

### Staff: Mentor

What do you mean by the fields pointing "outwards"? The fields are perpendicular to the direction of propagation.

3. May 5, 2013

### cryora

I mean outwards away from the Poynting vector instead of inwards towards the Poynting vector.

4. May 5, 2013

### Staff: Mentor

The Poynting vector points in the direction of propagation. The E&M fields are perpendicular to that direction.

5. May 5, 2013

### physwizard

In SI units,

$\bar{S}=\frac{1}{\mu_{0}}\bar{E}\times \bar{B}$

The vector S is the Poynting vector and points in the direction of flow of energy. E and B are the electric and magnetic fields. If you understand cross products you will not have any trouble relating the direction of the E and B fields to the direction of propagation.

6. May 5, 2013

### cryora

So normally we have the E-field vectors pointing away from the line along Poynting Vector.

I didn't draw the B-field, but you can imagine it's there.

What I'm wondering is if this is possible:

Mathematically speaking, even if either the E-field vectors or B-field vectors pointed towards the Poynting vector so that their heads, rather than their tails, touched, a cross product can still be performed to find the direction of the Poynting vector. I'm just wondering if there is any physical laws saying you can't orient them head-to-head.

7. May 5, 2013

### mathskier

Ah! So you're merely misunderstanding what that picture actually means. The wave is NOT a single ray, but rather a plane wave. That same picture is happening at each and every point in space along the same plane that is perpendicular to the direction of travel. Imagine those arrows translate all up and down the plane so that any point perpendicular to the direction of motion has the same vector as all other points along that plane.

Then there is no longer a sense of inward or outward, because there is not actually a line along which to define the concept.

8. May 5, 2013

### Staff: Mentor

The wavy lines in your pictures are defined by the direction and magnitude of the E-field vectors, so the second picture doesn't make any sense - you're using the wavy lines to position the vectors instead of the other way around.
The correct way of drawing the second picture is to just flip the first picture about the axis of travel so that the arrows on the left point down and the ones on the right point up.

Always remember that in pictures like these, the x-axis is a direction in space but the y-axis is not. These aren't drawings of waves, they're graphs of field strength as a function of position.

9. May 5, 2013

### cryora

@mathskier - Thanks, I never knew that! That makes a lot more sense. Although that makes me wonder, hypothetically speaking, what if we focused a beam of light so small that rather than a plane wave, it becomes a "point" wave, like a single photon, would we be able to use the same diagram to visualize the phenomenon?

@Nugatory - I see. So does that mean the electric field vectors only act along the line? What I previously thought was the electric field took up the entire area between the wave and line, acting on any charged particle that falls within that area.

Last edited: May 5, 2013
10. May 5, 2013

### Staff: Mentor

Yes, because that diagram already is a graph for the field strength at a single point. Pick a point on the x-axis, say x=X0. The y-axis of that graph gives you the field strength at the point x=X0, y=0, z=0.

11. May 5, 2013

### cryora

Got it. It makes a lot more sense now.

12. May 8, 2013

### cryora

I hope I'm not diverting this too much from Classical Physics, but knowing that electromagnetic waves propagate at speed c, and from experiments on the Photoelectric effect which establishes that the photon is the smallest quanta of EM radiation, would this picture be a good depiction of what a photon really is (not drawn perfectly to scale):

Basically it is a picture of a "photon" which carries an E-field and B-field at the point in which it is located, and the E-field and B-field oscillate, creating an EM wave due to the translational motion of the photon through space. Is this correct, or is it not this simple?

Another question is: Does the uncertainty principle apply to photons as well? In other words, in addition to being an electromagnetic wave, does the photon also have a "DeBroglie" wave property which prevents us from knowing both the momentum and position of the photon simultaneously?

I ask because from what I understand, interference of light comes from their electromagnetic wave properties, which has nothing to do with the wave property of electrons which cause electrons from interfering with each other. Photons would have to be located in the same position in order for their field strengths to add, so photons shouldn't interfere if we hypothetically fired one photon after another with a duration of time passing between each fired photon. I don't want to get two completely different things to be confused with each other.

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13. May 8, 2013

### vanhees71

No, this is an absolutely unjustified view about a photon. A photon is not a little billard ball like classical point particles. It's not even clear, what a photon's position might be. There is no classical picture which comes close to the correct QED description of a photon, which is by definition a normalizable one-particle Fock state.